If $T:V\to V$ linear and $N(V)+R(T)=V$ then $V=N(T)\oplus R(T)$

Question

If $T:V\to W$ linear and $N(V)+R(T)=V$ then $V=N(T)\oplus R(T)$ and $V$ is finite dimensional.

I would appreciate this in as much detail as possible. I have made several attempts at solutions trying to either get a contradiction or show directly which have led me nowhere and are too lengthy to show here. Ultimately I need to use the finite dimensionality and the sum properties somewhere but I am utterly lost.

This ultimately boils down to showing $N(T)\cap R(T)=\{0\}$ Please do this by finding a basis for both spaces.

Thanks!

Answer 1

First note

$$\dim(V)=\dim(N(T)+R(T))=\dim(N(T))+\dim(R(T))-\dim(N(T)\cap R(T))$$

however, we know by Rank-Nullity that $\dim(V)=\dim(N(T))+\dim(R(T))$, so we must have $\dim(N(T)\cap R(T))=0$. That gives the result.

Let me know if you haven't seen any of these theorems before and we can try to figure out a way without using them.

Answer 2

Suppose $x_1 \in \mathcal{N}(T)\cap\mathcal{R}(T)$ is non-zero, and further suppose that $X=\mathcal{N}(T)+\mathcal{R}(T)$. Then $x_1 \in \mathcal{R}(T)$ gives $x_1 = Tx_2 \ne 0$. Hence, $x_2 \ne 0$, $Tx_2 \ne 0$, $T^{2}x_2 =0$. Next write $$ x_2 = n_3+Tx_3,\;\;\; n_3 \in \mathcal{N}(T). $$ Then $x_3 \ne 0$ because $0\ne Tx_2 = Tn_3+T^{2}x_3 = T^{2}x_3$. So that gives $x_3 \ne 0, Tx_3 \ne 0, T^{2}x_3 \ne 0, T^{3}x_3=0$. Then, $$ x_3 = n_4+Tx_4,\;\;\; n_4 \in \mathcal{N}(T). $$ Then $x_4 \ne 0$ because $0 \ne Tx_3=Tn_4+T^{2}x_4=T^{2}x_4$. So that gives $x_4 \ne 0, Tx_4 \ne 0, T^{2}x_4 \ne 0, T^{3}x_4\ne 0, T^{4}x_4 = 0$. Keep going, and you get as long a chain as you want. However, such a chain of $k$ vectors $\{x_k,Tx_k,\cdots,T^{k-1}x_k\}$ is linearly independent if $$ x_k\ne 0,Tx_k\ne 0,T^{2}x_k\ne 0,\cdots,T^{k-1}x_k\ne 0,T^{k}x_k=0. $$ So, eventually, you reach a contradiction, once $k$ is greater than the dimension of the space.

Therefore, if $\mathcal{N}(T)+\mathcal{R}(T)=X$, then $\mathcal{N}(T)\cap\mathcal{R}(T)=\{0\}$, which gives $X=\mathcal{N}(T)\oplus\mathcal{R}(T)$.